Some Companions of Fejér's Inequality for Convex Functions

Some Companions of Fejér's Inequality for Convex Functions

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Tseng, Kuei-Lin, Hwang, Shiow-Ru and Dragomir, Sever S (2009) Some Companions of Fejér's Inequality for Convex Functions. Research report collection, 12 (Supp).

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SOME COMPANIONS OF FEJ ´ER’S INEQUALITY FOR CONVEX FUNCTIONS

K.-L. TSENG, SHIOW-RU HWANG, AND S.S. DRAGOMIR

Abstract. In this paper, we establish some companions of Fej´er’s inequality for convex functions which generalize the inequalities of Hermite-Hadamard type from [2] and [7].

1. Introduction

In what follows we assume that the functionf : [a, b]→Ris convex,g: [a, b]→ [0,∞) is integrable and symmetric to ^a+b₂ and we define the following associated functions on [0,1] by:

G(t) = 1 2

f

ta+ (1−t)a+b 2

+f

tb+ (1−t)a+b 2

; H(t) = 1

b−a Z b

a

f

tx+ (1−t)a+b 2

dx;

I(t) = Z b

a

1 2

f

tx+a

2 + (1−t)a+b 2

+f

tx+b

2 + (1−t)a+b 2

g(x)dx;

L(t) = 1 2 (b−a)

Z b

a

[f(ta+ (1−t)x) +f(tb+ (1−t)x)]dx;

and

Sg(t) =1 4

Z b

a

f

ta+ (1−t)x+a 2

+f

ta+ (1−t)x+b 2

+f

tb+ (1−t)x+a 2

+f

tb+ (1−t)x+b 2

g(x)dx.

Iff is defined as above, then

(1.1) f

a+b 2

≤ 1 b−a

Z b

a

f(x)dx≤ f(a) +f(b) 2 is known as the Hermite-Hadamard inequality [1].

For some results which generalize, improve, and extend this famous integral inequality see [2] – [16].

1991Mathematics Subject Classification. 26D15.

Key words and phrases. Hermite-Hadamard inequality, Fej´er inequality, Convex function.

This research was partially supported by grant NSC 98-2115-M-156-004.

1

In [2], Dragomir established the following theorem which refines the first inequal- ity of (1.1).

Theorem A. Letf, H be defined as above. ThenH is convex, increasing on[0,1], and for allt∈[0,1], we have

(1.2) f

a+b 2

=H(0)≤H(t)≤H(1) = 1 b−a

Z b

a

f(x)dx.

In [7], Dragomir, Miloˇsevi´c and S´andor established inequalities related to (1.1).

They are incorporated in the following:

Theorem B. Let f,H be defined as above. Then:

(1) The following inequality holds f

a+b 2

≤ 2 b−a

Z ^a+3b₄

3a+b 4

f(x)dx≤ Z 1

0

H(t)dt

≤ 1 2

"

f a+b

2

+ 1

b−a Z b

a

f(x)dx

# . (1.3)

(2) Iff is differentiable on[a, b],then, for allt∈[0,1], we have the inequalities 0≤ 1

b−a Z b

a

f(x)dx−H(t)

≤(1−t)

"

f(a) +f(b)

2 − 1

b−a Z b

a

f(x)dx

# (1.4)

and

(1.5) 0≤f(a) +f(b)

2 −H(t)≤ (f⁰(b)−f⁰(a)) (b−a)

4 .

Theorem C. Letf, H, Gbe defined as above. Then:

(1) Gis convex and increasing on[0,1].

(2) We have

inf

t∈[0,1]G(t) =G(0) =f a+b

2

and

sup

t∈[0,1]

G(t) =G(1) =f(a) +f(b)

2 .

(3) The following inequality holds for allt∈[0,1]:

(1.6) H(t)≤G(t).

(4) The following inequality holds:

2 b−a

Z ^a+3b₄

3a+b 4

f(x)dx≤ 1 2

f

3a+b 4

+f

a+ 3b 4

≤ Z 1

0

G(t)dt

≤ 1 2

f

a+b 2

+f(a) +f(b) 2

. (1.7)

(5) Iff is differentiable on[a, b],then, for allt∈[0,1],we have the inequality

(1.8) 0≤H(t)−f

a+b 2

≤G(t)−H(t). Theorem D. Let f, H, G, Lbe defined as above. Then:

(1) Lis convex on [0,1].

(2) We have the inequality:

(1.9) G(t)≤L(t)≤ 1−t b−a

Z b

a

f(x)dx+t·f(a) +f(b)

2 ≤ f(a) +f(b) 2 for allt∈[0,1]and

sup

t∈[0,1]

L(t) =f(a) +f(b)

2 .

(3) For allt∈[0,1], we have the inequalities:

H(1−t)≤L(t) and H(t) +H(1−t)

2 ≤L(t).

In [8], Fej´er established the following weighted generalization of the Hermite- Hadamard inequality (1.1).

Theorem E. Letf, g be defined as above. Then we have (1.10) f

a+b 2

Z b

a

g(x)dx≤ Z b

a

f(x)g(x)dx≤ f(a) +f(b) 2

Z b

a

g(x)dx, which is known as Fej´er’s inequality.

In [11], Tseng, Hwang and Dragomir established the following theorems related to Fej´er-type inequalities.

Theorem F. Let f, g, Ibe defined as above. ThenI is convex, increasing on[0,1], and for allt∈[0,1], we have the following Fej´er-type inequality

f a+b

2 Z b

a

g(x)dx=I(0)≤I(t)≤I(1)

= Z b

a

1 2

f

x+a 2

+f

x+b 2

g(x)dx.

(1.11)

Theorem G. Let f, gbe defined as above. Then we have f

a+b 2

Z b

a

g(x)dx≤ f ^3a+b₄

+f ^a+3b₄ 2

Z b

a

g(x)dx

≤ Z b

a

1 2

f

x+a 2

+f

x+b 2

g(x)dx

≤ 1 2

f

a+b 2

+f(a) +f(b) 2

Z b

a

g(x)dx

≤ f(a) +f(b) 2

Z b

a

g(x)dx.

(1.12)

In this paper, we establish other Fej´er-type inequalities related to the functions G, H, I, L, S_g and therefore generalize Theorems B – D from above.

2. Main Results

In order to prove our main results, we need the following simple lemma:

Lemma 1 (see [10]). Let f be defined as above and leta≤A≤C ≤D ≤B ≤b withA+B=C+D. Then

f(C) +f(D)≤f(A) +f(B). Now, we are ready to state and prove our results.

Theorem 2. Let f, g, I be defined as above. Then:

(1) The following inequality holds:

f a+b

2 Z b

a

g(x)dx

≤2

"

Z ^a+b₂

3a+b 4

f(x)g(4x−2a−b)dx+ Z ^a+3b₄

a+b 2

f(x)g(4x−a−2b)dx

#

≤ Z 1

0

I(t)dt

≤ 1 2

"

f a+b

2 Z b

a

g(x)dx (2.1)

+ Z b

a

1 2

f

x+a 2

+f

x+b 2

g(x)dx

# .

(2) If f is differentiable on [a, b] and g is bounded on [a, b], then, for all t ∈ [0,1], we have the inequality

0≤ Z b

a

1 2

f

x+a 2

+f

x+b 2

g(x)dx−I(t)

≤(1−t)

"

f(a) +f(b)

2 (b−a)− Z b

a

f(x)dx

# kgk_∞, (2.2)

wherekgk_∞= sup

x∈[a,b]

|g(x)|.

(3) If f is differentiable on[a, b], then, for allt∈[0,1],we have the inequality 0≤f(a) +f(b)

2

Z b

a

g(x)dx−I(t)

≤(f⁰(b)−f⁰(a)) (b−a) 4

Z b

a

g(x)dx.

(2.3)

Proof. (1) Using simple techniques of integration, under the hypothesis of g, we have the following identities:

(2.4) f a+b

2 Z b

a

g(x)dx= 4 Z ^a+b₂

a

Z ¹₂

0

f a+b

2

g(2x−a)dtdx;

2

"

Z ^a+b₂

3a+b 4

f(x)g(4x−2a−b)dx+ Z ^a+3b₄

a+b 2

f(x)g(4x−a−2b)dx

#

= 2 Z ^a+b₂

3a+b 4

[f(x) +f(a+b−x)]g(4x−2a−b)dx

= 2 Z ^a+b₂

a

Z ¹₂

0

f

x

2 +a+b 4

+f

3 (a+b)

4 −x

2

g(2x−a)dtdx;

(2.5)

Z 1

0

I(t)dt= Z b

a

Z ¹₂

0

1 2

f

tx+a

2 + (1−t)a+b 2

+f

(1−t)x+a

2 +ta+b 2

g(x)dtdx +

Z b

a

Z ¹₂

0

1 2

f

tx+b

2 + (1−t)a+b 2

+f

(1−t)x+b

2 +ta+b 2

g(x)dtdx

= Z b

a

Z ¹₂

0

1 2

f

tx+a

2 + (1−t)a+b 2

+f

(1−t)x+a

2 +ta+b 2

g(x)dtdx +

Z b

a

Z ¹₂

0

1 2

f

ta+ 2b−x

2 + (1−t)a+b 2

+f

(1−t)a+ 2b−x

2 +ta+b 2

g(x)dtdx

= Z ^a+b₂

a

Z ¹₂

0

f

ta+b

2 + (1−t)x (2.6)

+f

tx+ (1−t)a+b 2

g(2x−a)dtdx +

Z ^a+b₂

a

Z ¹₂

0

f

t(a+b−x) + (1−t)a+b 2

+f

ta+b

2 + (1−t) (a+b−x)

g(2x−a)dtdx;

1 2

"

f a+b

2 Z b

a

g(x)dx+ Z b

a

1 2

f

x+a 2

+f

x+b 2

g(x)dx

#

=1 2

Z b

a

Z ¹₂

0

2f

a+b 2

+f

x+a 2

+f

a+ 2b−x 2

g(x)dtdx

= Z ^a+b₂

a

Z ¹₂

0

f(x) +f a+b

2

g(2x−a)dtdx (2.7)

+ Z ^a+b₂

a

Z ¹₂

0

f

a+b 2

+f(a+b−x)

g(2x−a)dtdx.

By Lemma 1, the following inequalities hold for allt∈ 0,¹₂

andx∈ a,^a+b₂

.

(2.8) 4f

a+b 2

≤2

f x

2 +a+b 4

+f

3 (a+b)

4 −x

2

holds whenA= ^x₂ +^a+b₄ , C=D= ^a+b₂ andB =^3(a+b)₄ −^x₂ in Lemma 1.

(2.9) 2f x

2 +a+b 4

≤f

ta+b

2 + (1−t)x

+f

tx+ (1−t)a+b 2

holds whenA =t^a+b₂ + (1−t)x, C =D = ^x₂ +^a+b₄ and B =tx+ (1−t)^a+b₂ in Lemma 1.

(2.10) 2f

3 (a+b)

4 −x

2

≤f

t(a+b−x) + (1−t)a+b 2

+f

ta+b

2 + (1−t) (a+b−x)

holds whenA=t(a+b−x) + (1−t)^a+b₂ , C =D= ^3(a+b)₄ −^x₂ andB =t^a+b₂ + (1−t) (a+b−x) in Lemma 1.

(2.11) f

ta+b

2 + (1−t)x

+f

tx+ (1−t)a+b 2

≤f(x) +f a+b

2

holds when A =x, C =t^a+b₂ + (1−t)x, D =tx+ (1−t)^a+b₂ and B = ^a+b₂ in Lemma 1.

(2.12) f

t(a+b−x) + (1−t)a+b 2

+f

ta+b

2 + (1−t) (a+b−x)

≤f a+b

2

+f(a+b−x) holds for A= ^a+b₂ , C =t(a+b−x) + (1−t)^a+b₂ , D=t^a+b₂ + (1−t) (a+b−x) and B = a+b−x in Lemma 1. Multiplying the inequalities (2.8)−(2.12) by g(2x−a) and integrating them over t on

0,¹₂

, over x on a,^a+b₂

and using identities (2.4)−(2.7), we derive (2.1).

(2) On utilising the integration by parts, we have Z ^a+b₂

a

a+b 2 −x

[f⁰(a+b−x)−f⁰(x)]dx

= Z b

a

x−a+b 2

f⁰(x)dx

= f(a) +f(b)

2 (b−a)− Z b

a

f(x)dx.

(2.13)

Using substitution rules for integration, under the hypothesis of g, we have the following identities

Z b

a

1 2

f

x+a 2

+f

x+b 2

g(x)dx

= Z b

a

1 2

f

x+a 2

+f

a+ 2b−x 2

g(x)dx

= Z ^a+b₂

a

[f(x) +f(a+b−x)]g(2x−a)dx (2.14)

and

I(t) = Z b

a

1 2

f

tx+a

2 + (1−t)a+b 2

+f

ta+ 2b−x

2 + (1−t)a+b 2

g(x)dx

= Z ^a+b₂

a

f

tx+ (1−t)a+b 2

(2.15)

+f

t(a+b−x) + (1−t)a+b 2

g(2x−a)dx for allt∈[0,1].

Now, by the convexity off and the hypothesis ofg, the inequality

f(x)−f

tx+ (1−t)a+b 2

g(2x−a) +

f(a+b−x)−f

t(a+b−x) + (1−t)a+b 2

g(2x−a)

≤(1−t)

x−a+b 2

f⁰(x)g(2x−a) + (1−t)

a+b 2 −x

f⁰(a+b−x)g(2x−a)

= (1−t) a+b

2 −x

[f⁰(a+b−x)−f⁰(x)]g(2x−a)

≤(1−t) a+b

2 −x

[f⁰(a+b−x)−f⁰(x)]kgk_∞ holds for all t ∈[0,1] and x∈

a,^a+b₂

. Integrating the above inequalities overx on

a,^a+b₂

and using (2.13)−(2.15) and (1.11), we derive (2.2). (3) Using the convexity off, we have

f(a)−f ^a+b₂

2 ≤ 1

2

a−a+b 2

f⁰(a) = a−b 4 f⁰(a) and

f(b)−f ^a+b₂

2 ≤1

2

b−a+b 2

f⁰(b) = b−a 4 f⁰(b)

and taking their sum we obtain f(a) +f(b)

2 −f

a+b 2

≤ (f⁰(b)−f⁰(a)) (b−a)

4 .

Thus,

(2.16) f(a) +f(b) 2

Z b

a

g(x)dx−f a+b

2 Z b

a

g(x)dx

≤ (f⁰(b)−f⁰(a)) (b−a) 4

Z b

a

g(x)dx.

Finally, (2.3) follows from (1.11),(1.12) and (2.16). This completes the proof.

Remark 3. Letg(x) =_b−a¹ (x∈[a, b])in Theorem 2. ThenI(t) =H(t) (t∈[0,1]) and therefore Theorem 2 reduces to Theorem B.

In the following theorems, we shall point out some inequalities for the functions G, H, I, Sg considered above:

Theorem 4. Let f, g, G, I be defined as above. Then:

(1) The following inequality holds for allt∈[0,1]:

(2.17) I(t)≤G(t)

Z b

a

g(x)dx.

(2) If f is differentiable on [a, b] and g is bounded on [a, b], then, for all t ∈ [0,1], we have the inequality

(2.18) 0≤I(t)−f a+b

2 Z b

a

g(x)dx≤(b−a) [G(t)−H(t)]kgk_∞, wherekgk_∞= sup

x∈[a,b]

|g(x)|.

Proof. (1) Using simple techniques of integration, under the hypothesis of g, we have that the following identity holds on [0,1]:

(2.19) G(t) Z b

a

g(x)dx= Z ^a+b₂

a

f

ta+ (1−t)a+b 2

+f

tb+ (1−t)a+b 2

g(2x−a)dx.

By Lemma 1, the following inequality holds for allt∈[0,1] andx∈ a,^a+b₂

. (2.20) f

tx+ (1−t)a+b 2

+f

t(a+b−x) + (1−t)a+b 2

≤f

ta+ (1−t)a+b 2

+f

tb+ (1−t)a+b 2

holds whenA=ta+(1−t)^a+b₂ , C=tx+(1−t)^a+b₂ , D=t(a+b−x)+(1−t)^a+b₂ and B = tb+ (1−t)^a+b₂ in Lemma 1. Multiplying the inequality (2.20) by g(2x−a), integrating both sides over x on

a,^a+b₂

and using identities (2.15) and (2.19), we derive (2.17).

(2) Using an integration by parts, we have that the following identity holds on [0,1]:

t Z ^a+b₂

a

x−a+b 2

f⁰

tx+ (1−t)a+b 2

+ a+b

2 −x

f⁰

t(a+b−x) + (1−t)a+b 2

dx

=t Z b

a

x−a+b 2

f⁰

tx+ (1−t)a+b 2

dx

= (b−a) [G(t)−H(t)]. (2.21)

Now, by the convexity off,under the hypothesis ofg, the inequality

f

tx+ (1−t)a+b 2

−f a+b

2

g(2x−a) +

f

t(a+b−x) + (1−t)a+b 2

−f a+b

2

g(2x−a)

≤t

x−a+b 2

f⁰

tx+ (1−t)a+b 2

g(2x−a) +t

a+b 2 −x

f⁰

t(a+b−x) + (1−t)a+b 2

g(2x−a)

=t a+b

2 −x f⁰

t(a+b−x) + (1−t)a+b 2

−f⁰

tx+ (1−t)a+b 2

g(2x−a)

≤t a+b

2 −x f⁰

t(a+b−x) + (1−t)a+b 2

−f⁰

tx+ (1−t)a+b 2

kgk_∞

holds for allt∈[0,1] andx∈ a,^a+b₂

. Integrating the above inequality over xon a,^a+b₂

and using (2.21) and (1.11), we derive (2.18).This completes the proof.

Remark 5. Letg(x) =_b−a¹ (x∈[a, b])in Theorem 4. ThenI(t) =H(t) (t∈[0,1]) and the inequalities (2.17)and(2.18)reduce to the inequalities(1.6) and(1.8), re- spectively.

Theorem 6. Let f, g, G, I, Sg be defined as above. Then we have the following results:

(1) S_g is convex on [0,1].

(2) The following inequalities hold for allt∈[0,1]:

G(t) Z b

a

g(x)dx≤S_g(t)

≤(1−t) Z b

a

1 2

f

x+a 2

+f

x+b 2

g(x)dx +t·f(a) +f(b)

2

Z b

a

g(x)dx

≤f(a) +f(b) 2

Z b

a

g(x)dx;

(2.22)

(2.23) I(1−t)≤Sg(t) ;

(2.24) I(t) +I(1−t)

2 ≤Sg(t). (3) The following inequality holds:

(2.25) sup

t∈[0,1]

Sg(t) =f(a) +f(b) 2

Z b

a

g(x)dx.

Proof. (1) It is easily observed from the convexity off that Sg is convex on [0,1]. (2) As for (1) of Theorem 4, we have that the following identity holds on [0,1]:

(2.26) Sg(t) =1 2

Z ^a+b₂

a

[f(ta+ (1−t)x) +f(ta+ (1−t) (a+b−x))

+f(tb+ (1−t)x) +f(tb+ (1−t) (a+b−x))]g(2x−a)dx.

By Lemma 1, the following inequalities hold for allt∈[0,1] andx∈ a,^a+b₂

. (2.27) 2f

ta+ (1−t)a+b 2

≤f(ta+ (1−t)x) +f(ta+ (1−t) (a+b−x)) holds whenA=ta+(1−t)x, C=D=ta+(1−t)^a+b₂ andB=ta+(1−t) (a+b−x) in Lemma 1.

(2.28) 2f

tb+ (1−t)a+b 2

≤f(tb+ (1−t)x) +f(tb+ (1−t) (a+b−x)) holds whenA=tb+(1−t)x, C=D=tb+(1−t)^a+b₂ andB=tb+(1−t) (a+b−x) in Lemma 1. Multiplying the inequalities (2.27)−(2.28) byg(2x−a),integrating them overxon

a,^a+b₂

and using identities (2.19) and (2.26), we derive the first inequality of (2.22). Using the convexity of f and the inequality (1.12), the last

part of (2.22) holds. Next, by the convexity off and the identity (2.26), we get Ig(1−t) =

Z ^a+b₂

a

f

(1−t)x+ta+b 2

+f

(1−t) (a+b−x) +ta+b 2

g(2x−a)dx

= Z ^a+b₂

a

f

1

2(ta+ (1−t)x) +1

2(tb+ (1−t)x)

+f 1

2(ta+ (1−t) (a+b−x)) +1

2(tb+ (1−t) (a+b−x))

g(2x−a)dx

≤S_g(t) (2.29)

and the inequality (2.23) is proved. From (2.17), (2.22) and (2.23), we obtain (2.24).

(3) Using (2.22),the inequality (2.25) holds. This completes the proof.

Remark 7. Letg(x) =_b−a¹ (x∈[a, b])in Theorem 6. ThenI(t) =H(t) (t∈[0,1]), S_g(t) =L(t) (t∈[0,1]) and Theorem 6 reduces to Theorem D.

References

[1] J. Hadamard, ´Etude sur les propri´et´es des fonctions enti`eres en particulier d’une fonction consid´er´ee par Riemann,J. Math. Pures Appl.,58(1893), 171–215.

[2] S. S. Dragomir, Two mappings in connection to Hadamard’s inequalities, J. Math. Anal.

Appl.,167(1992), 49–56.

[3] S. S. Dragomir, A refinement of Hadamard’s inequality for isotonic linear functionals, Tamkang. J. Math.

[4] S. S. Dragomir, On the Hadamard’s inequality for convex on the co-ordinates in a rectangle from the plane,Taiwanese J. Math.,5(4) (2001), 775–788.

[5] S. S. Dragomir, Further properties of some mappings associated with Hermite-Hadamard inequalities,Tamkang. J. Math.,34(1) (2003), 45–57.

[6] S. S. Dragomir, Y. J. Cho and S. S. Kim, Inequalities of Hadamard’s type for Lipschitzian mappings and their applications,J. Math. Anal. Appl.,245(2000), 489–501.

[7] S. S. Dragomir, D.S. Miloˇsevi´c and J. S´andor, On some refinements of Hadamard’s inequalities and applications,Univ. Belgrad. Publ. Elek. Fak. Sci. Math.,4(1993), 3–10.

[8] L. Fej´er, ¨Uber die Fourierreihen, II,Math. Naturwiss.Anz Ungar. Akad. Wiss., 24(1906), 369–390. (Hungarian).

[9] D. Y. Hwang, K. L. Tseng and G. S. Yang, Some Hadamard’s inequalities for co-ordinated convex functions in a rectangle from the plane,Taiwanese J. Math.,11(1) (2007), 63–73.

[10] K. L. Tseng, S. R. Hwang and S. S. Dragomir, On some new inequalities of Hermite- Hadamard-Fej´er type involving convex functions,Demonstratio Math.,XL(1) (2007), 51–64.

[11] K. L. Tseng, S. R. Hwang and S. S. Dragomir, Fej´er-type inequalities (I). Preprint, RGMIA Res. Rep. Coll., 12(2009), No. 4, Article 5. [Online http://www.staff.vu.edu.au/RGMIA/v12n4.asp].

[12] G. S. Yang and M. C. Hong, A note on Hadamard’s inequality,Tamkang. J. Math.,28(1) (1997), 33–37.

[13] G. S. Yang and K. L. Tseng, On certain integral inequalities related to Hermite-Hadamard inequalities,J. Math. Anal. Appl.,239(1999), 180–187.

[14] G. S. Yang and K. L. Tseng, Inequalities of Hadamard’s type for Lipschitzian mappings,J.

Math. Anal. Appl.,260(2001), 230–238.

[15] G. S. Yang and K. L. Tseng, On certain multiple integral inequalities related to Hermite- Hadamard inequalities,Utilitas Math.,62(2002), 131–142.

[16] G. S. Yang and K. L. Tseng, Inequalities of Hermite-Hadamard-Fej´er type for convex functions and Lipschitzian functions,Taiwanese J. Math.,7(3) (2003), 433–440.

Department of Mathematics, Aletheia University, Tamsui, Taiwan 25103.

E-mail address:kltseng@email.au.edu.tw

China Institute of Technology, Nankang, Taipei, Taiwan 11522 E-mail address:shru@cc.chit.edu.tw

School of Engineering and Science, Victoria University, PO Box 14428, Melbourne City MC, Victoria 8001, Australia.

E-mail address:sever.dragomir@.vu.edu.au URL:http://www.staff.vu.edu.au/RGMIA/dragomir/